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Proof step 6 from proof gcd-one (previous step)

i0 | i ∧ i0 | j ∧ 0 ≠ i → i0 ≤ gcd(i, j), gcd(i, j) = 1, i0 | i, i0 | j, 0 < i0, 1 < i0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: div-one: ⊦ 1 | i;%gcd-zero-04: ⊦ gcd(i, 0) = abs(i);%gcd-com: ⊦ gcd(i, j) = gcd(j, i);%div-zero-01: ⊦ i | 0;%bet: ⊦ 0 < i → (i ≤ 1 ↔ i = 1); from specification int-abs